Cho a,b là các số thực dương, cmr: \(\left(\frac{a^2}{a}+\frac{b^2}{b}\right)>=2\left(a^2+b^2\right)\)
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Ta có \(VT=a^2+b^2+c^2+2\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\)
\(\Leftrightarrow VT=a^2+b^2+c^2+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(ab^2+bc^2+ca^2\right)\) (Vì abc=1)
ÁP dụng bđt Cô-si cho 3 số dương, ta có:\(a^2+\frac{1}{b^2}+ab^2\ge3\sqrt[3]{\frac{a^3b^2}{b^2}}=3a\)
\(b^2+\frac{1}{c^2}+bc^2\ge3b\) \(c^2+\frac{1}{a^2}+ca^2\ge3c\)
\(\Rightarrow VT\ge3\left(a+b+c\right)+\left(ab^2+bc^2+ca^2\right)\ge3\left(a+b+c\right)+3\sqrt[3]{a^3b^3c^3}=3\left(a+b+c+1\right)\) Vì abc=1. Dấu bằng xảy ra khi a=b=c=1
\(VT=\left(\sqrt{a^2}+\sqrt{b^2}+\sqrt{c^2}\right)\left[\left(\frac{\sqrt{a}}{b+c}\right)^2+\left(\frac{\sqrt{b}}{c+a}\right)^2+\left(\frac{\sqrt{c}}{a+b}\right)^2\right]\)
Áp dúng bất đẳng thức Bunhiacopxki ta có :
\(VT\ge\left(\sqrt{a}.\frac{\sqrt{a}}{b+c}+\sqrt{b}.\frac{\sqrt{b}}{c+a}+\sqrt{c}.\frac{\sqrt{c}}{a+b}\right)^2\)
\(\Leftrightarrow VT\ge\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\)
Xét \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Áp dụng bất đẳng thức Cauchy dạng phân thức ta có :
\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ca+bc}\)
\(\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}=\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ac\right)}=\frac{3}{2}\)
\(\Rightarrow\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\ge\left(\frac{3}{2}\right)^2=\frac{9}{4}\)
\(\Rightarrow VT\ge\frac{9}{4}\left(đpcm\right)\)
Dấu " = " xảy ra khi \(a=b=c\)
Chúc bạn học tốt !!!
\(\left(a+b\right)^2\le2\left(a^2+b^2\right)\)
=> BDT cần CMR <=> \(\frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\ge\frac{a^2}{a^2+b^2}+\frac{b^2}{b^2+c^2}+\frac{c^2}{c^2+a^2}\)
Ta có \(\frac{a^3}{a^2+b^2}=a-\frac{ab^2}{a^2+b^2}\ge a-\frac{ab^2}{2ab}=a-\frac{b}{2}\)
=>VT\(\ge\frac{a+b+c}{2}\) (Hơi tắt nên tự hiểu)
Ta đi Cm \(\frac{a+b+c}{2}\ge\frac{a^2}{a^2+b^2}+\frac{b^2}{b^2+c^2}+\frac{c^2}{c^2+a^2}\)
<=> \(\frac{a+b+c}{2}+\frac{b^2}{a^2+b^2}+\frac{c^2}{c^2+b^2}+\frac{a^2}{a^2+c^2}\ge3\)(*)
\(\frac{a+b+c}{2}\ge\frac{3}{2}\)
\(\frac{b^2}{a^2+b^2}+\frac{c^2}{c^2+b^2}+\frac{a^2}{c^2+a^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{2\left(a^2b^2+b^2c^2+c^2a^2\right)}\ge\frac{3}{2}\)
=>VT (*) \(\ge3\). Từ đó ta có dpcm
Kiêm đâu lắm bài bdt hay. Gửi link
\(\Leftrightarrow\frac{\left(b+c\right)^2+a^2-2a\left(b+c\right)}{\left(b+c\right)^2+a^2}+\frac{\left(a+c\right)^2+b^2-2b\left(a+c\right)}{\left(a+c\right)^2+b^2}+\frac{\left(b+a\right)^2+c^2-2c\left(a+b\right)}{\left(a+b\right)^2+c^2}\ge\frac{3}{5}\)
\(\Leftrightarrow3-2\left(\frac{a\left(b+c\right)}{\left(b+c\right)^2+a^2}+\frac{b\left(a+c\right)}{\left(a+c\right)^2+b^2}+\frac{c\left(a+b\right)}{\left(a+b\right)^2+c^2}\right)\ge\frac{3}{5}\)
\(\Leftrightarrow\frac{a\left(b+c\right)}{\left(b+c\right)^2+a^2}+\frac{b\left(a+c\right)}{\left(a+c\right)^2+b^2}+\frac{c\left(a+b\right)}{\left(a+b\right)^2+c^2}\le\frac{6}{5}\)
Chuẩn hóa \(a+b+c=3\) (hay đặt \(x=\frac{3a}{a+b+c};y=\frac{3b}{a+b+c};z=\frac{3c}{a+b+c}\))
BĐT cần chứng minh trở thành:
\(\frac{a\left(3-a\right)}{\left(3-a\right)^2+a^2}+\frac{b\left(3-b\right)}{\left(3-b\right)^2+b^2}+\frac{c\left(3-c\right)}{\left(3-c\right)^2+c^2}\le\frac{6}{5}\)
Ta có đánh giá: \(\frac{a\left(3-a\right)}{\left(3-a\right)^2+a^2}\le\frac{9a+1}{25}\) ; \(\forall a\in\left(0;3\right)\)
\(\Leftrightarrow\left(a-1\right)^2\left(2a+1\right)\ge0\) (luôn đúng)
Tương tự: \(\frac{b\left(3-b\right)}{\left(3-b\right)^2+b^2}\le\frac{9b+1}{25};\frac{c\left(3-c\right)}{\left(3-c\right)^2+c^2}\le\frac{9c+1}{25}\)
Cộng vế với vế: \(VT\le\frac{9\left(a+b+c\right)+3}{25}=\frac{30}{25}=\frac{6}{5}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Đặt \(a=\frac{1}{x};b=\frac{1}{y};c=\frac{1}{z}\Rightarrow xyz=1\)
Không khó để chứng minh \(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\ge x+y+z\)
\(VT=\Sigma\frac{y^2z}{x^2\left(1+2z\right)}=\Sigma\frac{\left(\frac{y^2}{x^2}\right)}{\frac{1+2z}{z}}\ge\frac{\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right)^2}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+6}\)
\(\ge\frac{\left(x+y+z\right)^2}{xy+yz+zx+6}\ge\frac{\left(x+y+z\right)^2}{\frac{\left(x+y+z\right)^2}{3}+6}\)
Đặt \(t=x+y+z\ge3\sqrt[3]{xyz}=3\).Cần chứng minh:
\(f\left(t\right)=\frac{t^2}{\frac{t^2}{3}+6}\ge1\Leftrightarrow\frac{2}{3}\left(t-3\right)\left(t+3\right)\ge0\)(đúng)
IS that true?
Làm xong em mới nhận ra không cần đổi biến:D
Ta có:
\(\frac{a}{b}+\frac{a}{b}+\frac{b}{c}\ge3\sqrt[3]{\frac{a^2}{bc}}=3\sqrt[3]{\frac{a^3}{abc}}=3a\)
Tương tự: \(\frac{b}{c}+\frac{b}{c}+\frac{c}{a}\ge3b;\frac{c}{a}+\frac{c}{a}+\frac{a}{b}\ge3c\)
Cộng theo vế 3 BĐT trên suy ra \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge a+b+c\)
Trở lại bài toán: \(VT=\Sigma_{cyc}\frac{\left(\frac{a^2}{b^2}\right)}{c+2}\ge\frac{\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)^2}{a+b+c+6}\ge\frac{\left(a+b+c\right)^2}{a+b+c+6}=\frac{t^2}{t+6}\)(với \(t=a+b+c\ge3\sqrt[3]{abc}=3\))
Cần chúng minh: \(\frac{t^2}{t+6}\ge1\Leftrightarrow t^2-t-6\ge0\Leftrightarrow\left(t-3\right)\left(t+2\right)\ge0\)(đúng)
Đặt \(a-b=x;b-c=y;c-a=z\)
\(\Rightarrow x+y+z=a-b+b-c+c-a=0\)
Lúc đó: \(B=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Mà \(x+y+z=0\Rightarrow2\left(x+y+z\right)=0\Rightarrow\frac{2\left(x+y+z\right)}{xyz}=0\)
\(\Rightarrow B=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2\left(x+y+z\right)}{xyz}\)
\(=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{yz}+\frac{2}{xz}+\frac{2}{xy}\)
\(=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
\(1+a^2=a^2+ab+bc+ca=\left(a+b\right)\left(c+a\right)\)
Tương tự, ta có: \(1+b^2=\left(a+b\right)\left(b+c\right)\)\(;\)\(1+c^2=\left(b+c\right)\left(c+a\right)\)
\(\Rightarrow\)\(\frac{2}{\sqrt{\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)}}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) ( do a, b, c dương )
\(\frac{a}{1+a^2}+\frac{b}{1+b^2}+\frac{c}{1+c^2}=\frac{a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\frac{2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
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